Question

Ring of mass 6 kg and radius 40 cm is revolving at the rate of 300 revolutions per minute. Its moment of inertia will be

Answer 1

I=MR²=6 × 40/100 × 40/100 kg m² = 0.96kg m²
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Answer 2

Given:

• The mass of the ring = 60 Kg
• The radius of the ring = 40 cm = 0.4 m
• Total revolutions per minute = 300

To Find:

• The moment of Inertia of the ring.

Solution:

The formula to find the moment of inertia is given by,

I = M×$R^2$ → {equation 1}

Where "M" is the mass and "R" is the radius.

On substituting the given values in equation 1 we get,

⇒ I = 6×(0.4×0.4) {multiplying the terms in the bracket}

⇒ I = 6×0.16 {multiplying the terms}

⇒ I = 0.96 Kg $m^2$

∴ The moment of Inertia of the ring = 0.96 Kg $m^2$